# ---
# title: 897. Increasing Order Search Tree
# id: problem897
# author: Indigo
# date: 2021-02-01
# difficulty: Easy
# categories: Tree, Depth-first Search, Recursion
# link: <https://leetcode.com/problems/increasing-order-search-tree/description/>
# hidden: true
# ---
# 
# Given the `root` of a binary search tree, rearrange the tree in **in-order**
# so that the leftmost node in the tree is now the root of the tree, and every
# node has no left child and only one right child.
# 
# 
# 
# **Example 1:**
# 
# ![](https://assets.leetcode.com/uploads/2020/11/17/ex1.jpg)
# 
#     
#     
#     Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
#     Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
#     
# 
# **Example 2:**
# 
# ![](https://assets.leetcode.com/uploads/2020/11/17/ex2.jpg)
# 
#     
#     
#     Input: root = [5,1,7]
#     Output: [1,null,5,null,7]
#     
# 
# 
# 
# **Constraints:**
# 
#   * The number of nodes in the given tree will be in the range `[1, 100]`.
#   * `0 <= Node.val <= 1000`
# 
# 
## @lc code=start
using LeetCode

function increasing_bst(root::TreeNode{Int})
    p = TreeNode(0)
    q = p
    function mid_tranv(root::TreeNode{Int})
        (root.left !== nothing) && mid_tranv(root.left)
        q.right = TreeNode(root.val)
        q = q.right
        (root.right !== nothing) && mid_tranv(root.right)
        nothing
    end
    mid_tranv(root)
    p.right
end
## @lc code=end
